giải
ta có (60-15.x^2).(2x+1)=0
=>\(\left[{}\begin{matrix}60-15.x^2=0\\2x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}15.x^2=60\\2x=-1\end{matrix}\right.=>\left[{}\begin{matrix}x^2=4\\x=\frac{-1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)
Vậy x = 2 hoặc -1/2
Ta có: \(\left(60-15x^2\right)\cdot\left(2x+1\right)=0\)
\(\Leftrightarrow15\left(4-x^2\right)\left(2x+1\right)=0\)
mà 15≠0
nên \(\left[{}\begin{matrix}4-x^2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=4\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2;\frac{-1}{2}\right\}\)
